\(\int \frac {(d+e x)^{11/2}}{(a d e+(c d^2+a e^2) x+c d e x^2)^2} \, dx\) [183]

Optimal result

Integrand size = 37, antiderivative size = 178 \[ \int \frac {(d+e x)^{11/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\frac {7 e \left (c d^2-a e^2\right )^2 \sqrt {d+e x}}{c^4 d^4}+\frac {7 e \left (c d^2-a e^2\right ) (d+e x)^{3/2}}{3 c^3 d^3}+\frac {7 e (d+e x)^{5/2}}{5 c^2 d^2}-\frac {(d+e x)^{7/2}}{c d (a e+c d x)}-\frac {7 e \left (c d^2-a e^2\right )^{5/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{c^{9/2} d^{9/2}} \] Output:

7*e*(-a*e^2+c*d^2)^2*(e*x+d)^(1/2)/c^4/d^4+7/3*e*(-a*e^2+c*d^2)*(e*x+d)^(3 
/2)/c^3/d^3+7/5*e*(e*x+d)^(5/2)/c^2/d^2-(e*x+d)^(7/2)/c/d/(c*d*x+a*e)-7*e* 
(-a*e^2+c*d^2)^(5/2)*arctanh(c^(1/2)*d^(1/2)*(e*x+d)^(1/2)/(-a*e^2+c*d^2)^ 
(1/2))/c^(9/2)/d^(9/2)
 

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.07 \[ \int \frac {(d+e x)^{11/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\frac {\sqrt {d+e x} \left (105 a^3 e^6-35 a^2 c d e^4 (7 d-2 e x)+7 a c^2 d^2 e^2 \left (23 d^2-24 d e x-2 e^2 x^2\right )+c^3 d^3 \left (-15 d^3+116 d^2 e x+32 d e^2 x^2+6 e^3 x^3\right )\right )}{15 c^4 d^4 (a e+c d x)}-\frac {7 e \left (-c d^2+a e^2\right )^{5/2} \arctan \left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {-c d^2+a e^2}}\right )}{c^{9/2} d^{9/2}} \] Input:

Integrate[(d + e*x)^(11/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2,x]
 

Output:

(Sqrt[d + e*x]*(105*a^3*e^6 - 35*a^2*c*d*e^4*(7*d - 2*e*x) + 7*a*c^2*d^2*e 
^2*(23*d^2 - 24*d*e*x - 2*e^2*x^2) + c^3*d^3*(-15*d^3 + 116*d^2*e*x + 32*d 
*e^2*x^2 + 6*e^3*x^3)))/(15*c^4*d^4*(a*e + c*d*x)) - (7*e*(-(c*d^2) + a*e^ 
2)^(5/2)*ArcTan[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[-(c*d^2) + a*e^2]])/( 
c^(9/2)*d^(9/2))
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.17, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.189, Rules used = {1121, 51, 60, 60, 60, 73, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {(d+e x)^{11/2}}{\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^2} \, dx\)

\(\Big \downarrow \) 1121

\(\displaystyle \int \frac {(d+e x)^{7/2}}{(a e+c d x)^2}dx\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {7 e \int \frac {(d+e x)^{5/2}}{a e+c d x}dx}{2 c d}-\frac {(d+e x)^{7/2}}{c d (a e+c d x)}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {7 e \left (\frac {\left (d^2-\frac {a e^2}{c}\right ) \int \frac {(d+e x)^{3/2}}{a e+c d x}dx}{d}+\frac {2 (d+e x)^{5/2}}{5 c d}\right )}{2 c d}-\frac {(d+e x)^{7/2}}{c d (a e+c d x)}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {7 e \left (\frac {\left (d^2-\frac {a e^2}{c}\right ) \left (\frac {\left (d^2-\frac {a e^2}{c}\right ) \int \frac {\sqrt {d+e x}}{a e+c d x}dx}{d}+\frac {2 (d+e x)^{3/2}}{3 c d}\right )}{d}+\frac {2 (d+e x)^{5/2}}{5 c d}\right )}{2 c d}-\frac {(d+e x)^{7/2}}{c d (a e+c d x)}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {7 e \left (\frac {\left (d^2-\frac {a e^2}{c}\right ) \left (\frac {\left (d^2-\frac {a e^2}{c}\right ) \left (\frac {\left (d^2-\frac {a e^2}{c}\right ) \int \frac {1}{(a e+c d x) \sqrt {d+e x}}dx}{d}+\frac {2 \sqrt {d+e x}}{c d}\right )}{d}+\frac {2 (d+e x)^{3/2}}{3 c d}\right )}{d}+\frac {2 (d+e x)^{5/2}}{5 c d}\right )}{2 c d}-\frac {(d+e x)^{7/2}}{c d (a e+c d x)}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {7 e \left (\frac {\left (d^2-\frac {a e^2}{c}\right ) \left (\frac {\left (d^2-\frac {a e^2}{c}\right ) \left (\frac {2 \left (d^2-\frac {a e^2}{c}\right ) \int \frac {1}{-\frac {c d^2}{e}+\frac {c (d+e x) d}{e}+a e}d\sqrt {d+e x}}{d e}+\frac {2 \sqrt {d+e x}}{c d}\right )}{d}+\frac {2 (d+e x)^{3/2}}{3 c d}\right )}{d}+\frac {2 (d+e x)^{5/2}}{5 c d}\right )}{2 c d}-\frac {(d+e x)^{7/2}}{c d (a e+c d x)}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {7 e \left (\frac {\left (d^2-\frac {a e^2}{c}\right ) \left (\frac {\left (d^2-\frac {a e^2}{c}\right ) \left (\frac {2 \sqrt {d+e x}}{c d}-\frac {2 \left (d^2-\frac {a e^2}{c}\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\sqrt {c} d^{3/2} \sqrt {c d^2-a e^2}}\right )}{d}+\frac {2 (d+e x)^{3/2}}{3 c d}\right )}{d}+\frac {2 (d+e x)^{5/2}}{5 c d}\right )}{2 c d}-\frac {(d+e x)^{7/2}}{c d (a e+c d x)}\)

Input:

Int[(d + e*x)^(11/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2,x]
 

Output:

-((d + e*x)^(7/2)/(c*d*(a*e + c*d*x))) + (7*e*((2*(d + e*x)^(5/2))/(5*c*d) 
 + ((d^2 - (a*e^2)/c)*((2*(d + e*x)^(3/2))/(3*c*d) + ((d^2 - (a*e^2)/c)*(( 
2*Sqrt[d + e*x])/(c*d) - (2*(d^2 - (a*e^2)/c)*ArcTanh[(Sqrt[c]*Sqrt[d]*Sqr 
t[d + e*x])/Sqrt[c*d^2 - a*e^2]])/(Sqrt[c]*d^(3/2)*Sqrt[c*d^2 - a*e^2])))/ 
d))/d))/(2*c*d)
 

Defintions of rubi rules used

rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
 

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 1121

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_ 
Symbol] :> Int[ExpandIntegrand[(d + e*x)^(m + p)*(a/d + (c/e)*x)^p, x], x] 
/; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && (Int 
egerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && LtQ[c, 0]))
 

Maple [A] (verified)

Time = 4.12 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.15

method	result	size
pseudoelliptic	\(-\frac {7 \left (e \left (a \,e^{2}-c \,d^{2}\right )^{3} \left (c d x +a e \right ) \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )-\left (-\frac {\left (-\frac {2}{5} e^{3} x^{3}-\frac {32}{15} d \,e^{2} x^{2}-\frac {116}{15} d^{2} e x +d^{3}\right ) d^{3} c^{3}}{7}+\frac {23 e^{2} \left (-\frac {2}{23} e^{2} x^{2}-\frac {24}{23} d e x +d^{2}\right ) a \,d^{2} c^{2}}{15}-\frac {7 e^{4} \left (-\frac {2 e x}{7}+d \right ) a^{2} d c}{3}+e^{6} a^{3}\right ) \sqrt {e x +d}\, \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}\right )}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}\, d^{4} c^{4} \left (c d x +a e \right )}\)	\(204\)
risch	\(\frac {2 e \left (3 x^{2} c^{2} d^{2} e^{2}-10 x a c d \,e^{3}+16 x \,c^{2} d^{3} e +45 a^{2} e^{4}-100 a c \,d^{2} e^{2}+58 c^{2} d^{4}\right ) \sqrt {e x +d}}{15 d^{4} c^{4}}-\frac {\left (2 e^{6} a^{3}-6 d^{2} e^{4} a^{2} c +6 d^{4} e^{2} a \,c^{2}-2 d^{6} c^{3}\right ) e \left (-\frac {\sqrt {e x +d}}{2 \left (c d \left (e x +d \right )+a \,e^{2}-c \,d^{2}\right )}+\frac {7 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{2 \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{c^{4} d^{4}}\)	\(207\)
derivativedivides	\(2 e \left (\frac {\frac {c^{2} d^{2} \left (e x +d \right )^{\frac {5}{2}}}{5}-\frac {2 a c d \,e^{2} \left (e x +d \right )^{\frac {3}{2}}}{3}+\frac {2 c^{2} d^{3} \left (e x +d \right )^{\frac {3}{2}}}{3}+3 a^{2} e^{4} \sqrt {e x +d}-6 a c \,d^{2} e^{2} \sqrt {e x +d}+3 c^{2} d^{4} \sqrt {e x +d}}{c^{4} d^{4}}-\frac {\frac {\left (-\frac {1}{2} e^{6} a^{3}+\frac {3}{2} d^{2} e^{4} a^{2} c -\frac {3}{2} d^{4} e^{2} a \,c^{2}+\frac {1}{2} d^{6} c^{3}\right ) \sqrt {e x +d}}{c d \left (e x +d \right )+a \,e^{2}-c \,d^{2}}+\frac {7 \left (e^{6} a^{3}-3 d^{2} e^{4} a^{2} c +3 d^{4} e^{2} a \,c^{2}-d^{6} c^{3}\right ) \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{2 \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}}{c^{4} d^{4}}\right )\)	\(272\)
default	\(2 e \left (\frac {\frac {c^{2} d^{2} \left (e x +d \right )^{\frac {5}{2}}}{5}-\frac {2 a c d \,e^{2} \left (e x +d \right )^{\frac {3}{2}}}{3}+\frac {2 c^{2} d^{3} \left (e x +d \right )^{\frac {3}{2}}}{3}+3 a^{2} e^{4} \sqrt {e x +d}-6 a c \,d^{2} e^{2} \sqrt {e x +d}+3 c^{2} d^{4} \sqrt {e x +d}}{c^{4} d^{4}}-\frac {\frac {\left (-\frac {1}{2} e^{6} a^{3}+\frac {3}{2} d^{2} e^{4} a^{2} c -\frac {3}{2} d^{4} e^{2} a \,c^{2}+\frac {1}{2} d^{6} c^{3}\right ) \sqrt {e x +d}}{c d \left (e x +d \right )+a \,e^{2}-c \,d^{2}}+\frac {7 \left (e^{6} a^{3}-3 d^{2} e^{4} a^{2} c +3 d^{4} e^{2} a \,c^{2}-d^{6} c^{3}\right ) \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{2 \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}}{c^{4} d^{4}}\right )\)	\(272\)

Input:

int((e*x+d)^(11/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^2,x,method=_RETURNVER 
BOSE)
 

Output:

-7/(c*d*(a*e^2-c*d^2))^(1/2)*(e*(a*e^2-c*d^2)^3*(c*d*x+a*e)*arctan(c*d*(e* 
x+d)^(1/2)/(c*d*(a*e^2-c*d^2))^(1/2))-(-1/7*(-2/5*e^3*x^3-32/15*d*e^2*x^2- 
116/15*d^2*e*x+d^3)*d^3*c^3+23/15*e^2*(-2/23*e^2*x^2-24/23*d*e*x+d^2)*a*d^ 
2*c^2-7/3*e^4*(-2/7*e*x+d)*a^2*d*c+e^6*a^3)*(e*x+d)^(1/2)*(c*d*(a*e^2-c*d^ 
2))^(1/2))/d^4/c^4/(c*d*x+a*e)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 586, normalized size of antiderivative = 3.29 \[ \int \frac {(d+e x)^{11/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\left [\frac {105 \, {\left (a c^{2} d^{4} e^{2} - 2 \, a^{2} c d^{2} e^{4} + a^{3} e^{6} + {\left (c^{3} d^{5} e - 2 \, a c^{2} d^{3} e^{3} + a^{2} c d e^{5}\right )} x\right )} \sqrt {\frac {c d^{2} - a e^{2}}{c d}} \log \left (\frac {c d e x + 2 \, c d^{2} - a e^{2} - 2 \, \sqrt {e x + d} c d \sqrt {\frac {c d^{2} - a e^{2}}{c d}}}{c d x + a e}\right ) + 2 \, {\left (6 \, c^{3} d^{3} e^{3} x^{3} - 15 \, c^{3} d^{6} + 161 \, a c^{2} d^{4} e^{2} - 245 \, a^{2} c d^{2} e^{4} + 105 \, a^{3} e^{6} + 2 \, {\left (16 \, c^{3} d^{4} e^{2} - 7 \, a c^{2} d^{2} e^{4}\right )} x^{2} + 2 \, {\left (58 \, c^{3} d^{5} e - 84 \, a c^{2} d^{3} e^{3} + 35 \, a^{2} c d e^{5}\right )} x\right )} \sqrt {e x + d}}{30 \, {\left (c^{5} d^{5} x + a c^{4} d^{4} e\right )}}, -\frac {105 \, {\left (a c^{2} d^{4} e^{2} - 2 \, a^{2} c d^{2} e^{4} + a^{3} e^{6} + {\left (c^{3} d^{5} e - 2 \, a c^{2} d^{3} e^{3} + a^{2} c d e^{5}\right )} x\right )} \sqrt {-\frac {c d^{2} - a e^{2}}{c d}} \arctan \left (-\frac {\sqrt {e x + d} c d \sqrt {-\frac {c d^{2} - a e^{2}}{c d}}}{c d^{2} - a e^{2}}\right ) - {\left (6 \, c^{3} d^{3} e^{3} x^{3} - 15 \, c^{3} d^{6} + 161 \, a c^{2} d^{4} e^{2} - 245 \, a^{2} c d^{2} e^{4} + 105 \, a^{3} e^{6} + 2 \, {\left (16 \, c^{3} d^{4} e^{2} - 7 \, a c^{2} d^{2} e^{4}\right )} x^{2} + 2 \, {\left (58 \, c^{3} d^{5} e - 84 \, a c^{2} d^{3} e^{3} + 35 \, a^{2} c d e^{5}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (c^{5} d^{5} x + a c^{4} d^{4} e\right )}}\right ] \] Input:

integrate((e*x+d)^(11/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x, algorithm= 
"fricas")
 

Output:

[1/30*(105*(a*c^2*d^4*e^2 - 2*a^2*c*d^2*e^4 + a^3*e^6 + (c^3*d^5*e - 2*a*c 
^2*d^3*e^3 + a^2*c*d*e^5)*x)*sqrt((c*d^2 - a*e^2)/(c*d))*log((c*d*e*x + 2* 
c*d^2 - a*e^2 - 2*sqrt(e*x + d)*c*d*sqrt((c*d^2 - a*e^2)/(c*d)))/(c*d*x + 
a*e)) + 2*(6*c^3*d^3*e^3*x^3 - 15*c^3*d^6 + 161*a*c^2*d^4*e^2 - 245*a^2*c* 
d^2*e^4 + 105*a^3*e^6 + 2*(16*c^3*d^4*e^2 - 7*a*c^2*d^2*e^4)*x^2 + 2*(58*c 
^3*d^5*e - 84*a*c^2*d^3*e^3 + 35*a^2*c*d*e^5)*x)*sqrt(e*x + d))/(c^5*d^5*x 
 + a*c^4*d^4*e), -1/15*(105*(a*c^2*d^4*e^2 - 2*a^2*c*d^2*e^4 + a^3*e^6 + ( 
c^3*d^5*e - 2*a*c^2*d^3*e^3 + a^2*c*d*e^5)*x)*sqrt(-(c*d^2 - a*e^2)/(c*d)) 
*arctan(-sqrt(e*x + d)*c*d*sqrt(-(c*d^2 - a*e^2)/(c*d))/(c*d^2 - a*e^2)) - 
 (6*c^3*d^3*e^3*x^3 - 15*c^3*d^6 + 161*a*c^2*d^4*e^2 - 245*a^2*c*d^2*e^4 + 
 105*a^3*e^6 + 2*(16*c^3*d^4*e^2 - 7*a*c^2*d^2*e^4)*x^2 + 2*(58*c^3*d^5*e 
- 84*a*c^2*d^3*e^3 + 35*a^2*c*d*e^5)*x)*sqrt(e*x + d))/(c^5*d^5*x + a*c^4* 
d^4*e)]
                                                                                    
                                                                                    
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(d+e x)^{11/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\text {Timed out} \] Input:

integrate((e*x+d)**(11/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**2,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(d+e x)^{11/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((e*x+d)^(11/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x, algorithm= 
"maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(a*e^2-c*d^2>0)', see `assume?` f 
or more de
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 312 vs. \(2 (152) = 304\).

Time = 0.16 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.75 \[ \int \frac {(d+e x)^{11/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\frac {7 \, {\left (c^{3} d^{6} e - 3 \, a c^{2} d^{4} e^{3} + 3 \, a^{2} c d^{2} e^{5} - a^{3} e^{7}\right )} \arctan \left (\frac {\sqrt {e x + d} c d}{\sqrt {-c^{2} d^{3} + a c d e^{2}}}\right )}{\sqrt {-c^{2} d^{3} + a c d e^{2}} c^{4} d^{4}} - \frac {\sqrt {e x + d} c^{3} d^{6} e - 3 \, \sqrt {e x + d} a c^{2} d^{4} e^{3} + 3 \, \sqrt {e x + d} a^{2} c d^{2} e^{5} - \sqrt {e x + d} a^{3} e^{7}}{{\left ({\left (e x + d\right )} c d - c d^{2} + a e^{2}\right )} c^{4} d^{4}} + \frac {2 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} c^{8} d^{8} e + 10 \, {\left (e x + d\right )}^{\frac {3}{2}} c^{8} d^{9} e + 45 \, \sqrt {e x + d} c^{8} d^{10} e - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} a c^{7} d^{7} e^{3} - 90 \, \sqrt {e x + d} a c^{7} d^{8} e^{3} + 45 \, \sqrt {e x + d} a^{2} c^{6} d^{6} e^{5}\right )}}{15 \, c^{10} d^{10}} \] Input:

integrate((e*x+d)^(11/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x, algorithm= 
"giac")
 

Output:

7*(c^3*d^6*e - 3*a*c^2*d^4*e^3 + 3*a^2*c*d^2*e^5 - a^3*e^7)*arctan(sqrt(e* 
x + d)*c*d/sqrt(-c^2*d^3 + a*c*d*e^2))/(sqrt(-c^2*d^3 + a*c*d*e^2)*c^4*d^4 
) - (sqrt(e*x + d)*c^3*d^6*e - 3*sqrt(e*x + d)*a*c^2*d^4*e^3 + 3*sqrt(e*x 
+ d)*a^2*c*d^2*e^5 - sqrt(e*x + d)*a^3*e^7)/(((e*x + d)*c*d - c*d^2 + a*e^ 
2)*c^4*d^4) + 2/15*(3*(e*x + d)^(5/2)*c^8*d^8*e + 10*(e*x + d)^(3/2)*c^8*d 
^9*e + 45*sqrt(e*x + d)*c^8*d^10*e - 10*(e*x + d)^(3/2)*a*c^7*d^7*e^3 - 90 
*sqrt(e*x + d)*a*c^7*d^8*e^3 + 45*sqrt(e*x + d)*a^2*c^6*d^6*e^5)/(c^10*d^1 
0)
 

Mupad [B] (verification not implemented)

Time = 5.45 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.63 \[ \int \frac {(d+e x)^{11/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\frac {\sqrt {d+e\,x}\,\left (a^3\,e^7-3\,a^2\,c\,d^2\,e^5+3\,a\,c^2\,d^4\,e^3-c^3\,d^6\,e\right )}{c^5\,d^5\,\left (d+e\,x\right )-c^5\,d^6+a\,c^4\,d^4\,e^2}-\left (\frac {2\,e\,{\left (a\,e^2-c\,d^2\right )}^2}{c^4\,d^4}-\frac {2\,e\,{\left (2\,c^2\,d^3-2\,a\,c\,d\,e^2\right )}^2}{c^6\,d^6}\right )\,\sqrt {d+e\,x}+\frac {2\,e\,{\left (d+e\,x\right )}^{5/2}}{5\,c^2\,d^2}+\frac {2\,e\,\left (2\,c^2\,d^3-2\,a\,c\,d\,e^2\right )\,{\left (d+e\,x\right )}^{3/2}}{3\,c^4\,d^4}-\frac {7\,e\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,e\,{\left (a\,e^2-c\,d^2\right )}^{5/2}\,\sqrt {d+e\,x}}{a^3\,e^7-3\,a^2\,c\,d^2\,e^5+3\,a\,c^2\,d^4\,e^3-c^3\,d^6\,e}\right )\,{\left (a\,e^2-c\,d^2\right )}^{5/2}}{c^{9/2}\,d^{9/2}} \] Input:

int((d + e*x)^(11/2)/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^2,x)
 

Output:

((d + e*x)^(1/2)*(a^3*e^7 - c^3*d^6*e + 3*a*c^2*d^4*e^3 - 3*a^2*c*d^2*e^5) 
)/(c^5*d^5*(d + e*x) - c^5*d^6 + a*c^4*d^4*e^2) - ((2*e*(a*e^2 - c*d^2)^2) 
/(c^4*d^4) - (2*e*(2*c^2*d^3 - 2*a*c*d*e^2)^2)/(c^6*d^6))*(d + e*x)^(1/2) 
+ (2*e*(d + e*x)^(5/2))/(5*c^2*d^2) + (2*e*(2*c^2*d^3 - 2*a*c*d*e^2)*(d + 
e*x)^(3/2))/(3*c^4*d^4) - (7*e*atan((c^(1/2)*d^(1/2)*e*(a*e^2 - c*d^2)^(5/ 
2)*(d + e*x)^(1/2))/(a^3*e^7 - c^3*d^6*e + 3*a*c^2*d^4*e^3 - 3*a^2*c*d^2*e 
^5))*(a*e^2 - c*d^2)^(5/2))/(c^(9/2)*d^(9/2))
 

Reduce [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 570, normalized size of antiderivative = 3.20 \[ \int \frac {(d+e x)^{11/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\frac {-105 \sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}}\right ) a^{3} e^{6}+210 \sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}}\right ) a^{2} c \,d^{2} e^{4}-105 \sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}}\right ) a^{2} c d \,e^{5} x -105 \sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}}\right ) a \,c^{2} d^{4} e^{2}+210 \sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}}\right ) a \,c^{2} d^{3} e^{3} x -105 \sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}}\right ) c^{3} d^{5} e x +105 \sqrt {e x +d}\, a^{3} c d \,e^{6}-245 \sqrt {e x +d}\, a^{2} c^{2} d^{3} e^{4}+70 \sqrt {e x +d}\, a^{2} c^{2} d^{2} e^{5} x +161 \sqrt {e x +d}\, a \,c^{3} d^{5} e^{2}-168 \sqrt {e x +d}\, a \,c^{3} d^{4} e^{3} x -14 \sqrt {e x +d}\, a \,c^{3} d^{3} e^{4} x^{2}-15 \sqrt {e x +d}\, c^{4} d^{7}+116 \sqrt {e x +d}\, c^{4} d^{6} e x +32 \sqrt {e x +d}\, c^{4} d^{5} e^{2} x^{2}+6 \sqrt {e x +d}\, c^{4} d^{4} e^{3} x^{3}}{15 c^{5} d^{5} \left (c d x +a e \right )} \] Input:

int((e*x+d)^(11/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x)
 

Output:

( - 105*sqrt(d)*sqrt(c)*sqrt(a*e**2 - c*d**2)*atan((sqrt(d + e*x)*c*d)/(sq 
rt(d)*sqrt(c)*sqrt(a*e**2 - c*d**2)))*a**3*e**6 + 210*sqrt(d)*sqrt(c)*sqrt 
(a*e**2 - c*d**2)*atan((sqrt(d + e*x)*c*d)/(sqrt(d)*sqrt(c)*sqrt(a*e**2 - 
c*d**2)))*a**2*c*d**2*e**4 - 105*sqrt(d)*sqrt(c)*sqrt(a*e**2 - c*d**2)*ata 
n((sqrt(d + e*x)*c*d)/(sqrt(d)*sqrt(c)*sqrt(a*e**2 - c*d**2)))*a**2*c*d*e* 
*5*x - 105*sqrt(d)*sqrt(c)*sqrt(a*e**2 - c*d**2)*atan((sqrt(d + e*x)*c*d)/ 
(sqrt(d)*sqrt(c)*sqrt(a*e**2 - c*d**2)))*a*c**2*d**4*e**2 + 210*sqrt(d)*sq 
rt(c)*sqrt(a*e**2 - c*d**2)*atan((sqrt(d + e*x)*c*d)/(sqrt(d)*sqrt(c)*sqrt 
(a*e**2 - c*d**2)))*a*c**2*d**3*e**3*x - 105*sqrt(d)*sqrt(c)*sqrt(a*e**2 - 
 c*d**2)*atan((sqrt(d + e*x)*c*d)/(sqrt(d)*sqrt(c)*sqrt(a*e**2 - c*d**2))) 
*c**3*d**5*e*x + 105*sqrt(d + e*x)*a**3*c*d*e**6 - 245*sqrt(d + e*x)*a**2* 
c**2*d**3*e**4 + 70*sqrt(d + e*x)*a**2*c**2*d**2*e**5*x + 161*sqrt(d + e*x 
)*a*c**3*d**5*e**2 - 168*sqrt(d + e*x)*a*c**3*d**4*e**3*x - 14*sqrt(d + e* 
x)*a*c**3*d**3*e**4*x**2 - 15*sqrt(d + e*x)*c**4*d**7 + 116*sqrt(d + e*x)* 
c**4*d**6*e*x + 32*sqrt(d + e*x)*c**4*d**5*e**2*x**2 + 6*sqrt(d + e*x)*c** 
4*d**4*e**3*x**3)/(15*c**5*d**5*(a*e + c*d*x))